v1 · padrão canônico

Lição 40 — Workshop: Integral e Síntese de Cálculo 1

Workshop final de Cálculo 1: exercícios integradores cobrindo somas de Riemann, TFC, substituição, área, volume, comprimento de arco e aplicações físicas da integral.

Used in: Cálculo 1 — Unidade 4 · USP MAC0105 · ITA MA-011

\int_a^b f(x)\,dx = F(b) - F(a) \qquad \frac{d}{dx}\int_a^x f(t)\,dt = f(x)

Choose your door

Rigorous notation, full derivation, hypotheses

No content for this age yet.

Exercícios

Exercise list

33 exercises · 8 with worked solution (25%)

Application 21Understanding 8Modeling 2Challenge 2

Ex. 40.1Understanding
Expresse o limite como integral definida sobre o intervalo indicado:
$\lim_{n\to\infty}\sum_{i=1}^n x_i^*\,\Delta x \quad \text{sobre } [1,3]$
Select the correct option
$\displaystyle\int_1^3 x\,dx$ $\displaystyle\int_0^3 x\,dx$ $\displaystyle\int_1^3 x^2\,dx$ $\displaystyle\int_0^1 x\,dx$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 60
Show solution
A soma de Riemann $\lim_{n\to\infty}\sum_{i=1}^n x_i^*\Delta x$ sobre $[1,3]$ converge para a integral definida $\int_1^3 x\,dx$ .
Ex. 40.2Application
Expresse o limite como integral definida identificando o intervalo correto:
$R_n = \frac{1}{n}\sum_{i=1}^n \frac{i}{n}$
Select the correct option
$\displaystyle\int_0^1 x\,dx$ $\displaystyle\int_0^1 x^2\,dx$ $\displaystyle\int_0^1 \sqrt{x}\,dx$ $\displaystyle\int_1^2 x\,dx$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 65
Show solution
$R_n = \frac{1}{n}\sum_{i=1}^n \frac{i}{n}$ . Fazendo $x_i = i/n$ e $\Delta x = 1/n$ , o intervalo é $[0,1]$ e o integrando é $f(x) = x$ , logo $\int_0^1 x\,dx$ .
Show step-by-step (with the why)
1. Identifique: $\Delta x = 1/n$ e $x_i^* = i/n$ .
2. O intervalo vai de $x_0 = 0$ a $x_n = n/n = 1$ .
3. O integrando é $f(x) = x$ .
4. Portanto o limite é $\int_0^1 x\,dx = 1/2$ .
Ex. 40.3Application
Calcule usando fórmulas de área geométrica:
$\int_0^3 (3-x)\,dx$
Select the correct option
$\dfrac{9}{2}$ $3$ $\dfrac{3}{2}$ $6$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 76
Show solution
$\int_0^3(3-x)\,dx$ . O integrando é positivo em $[0,3]$ e forma um triângulo de base 3 e altura 3, logo área $= \tfrac{1}{2}\cdot3\cdot3 = \tfrac{9}{2}$ .
Ex. 40.4Application
Calcule usando fórmulas de área geométrica:
$\int_{-2}^2 \sqrt{4-x^2}\,dx$
Select the correct option
$2\pi$ $\pi$ $4\pi$ $\dfrac{\pi}{2}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 80
Show solution
$\int_{-2}^2\sqrt{4-x^2}\,dx$ é a área do semicírculo de raio 2, logo $\tfrac{1}{2}\pi r^2 = \tfrac{1}{2}\pi(4) = 2\pi$ .
Ex. 40.5Understanding
Dados $\int_0^4 f = 5$ , $\int_0^2 f = -3$ , $\int_0^4 g = -1$ , $\int_0^2 g = 2$ , calcule:
$\int_0^4 (f(x)+g(x))\,dx$
Select the correct option
$4$ $6$ $1$ $-4$
Select an option first
OpenStax Calculus Volume 1 · §5.2 · ex. 88
Show solution
Por aditividade: $\int_0^4(f+g) = \int_0^4 f + \int_0^4 g = 5 + (-1) = 4$ .
Ex. 40.6Understanding
Dados $\int_0^4 f = 5$ , $\int_0^2 f = -3$ , $\int_0^4 g = -1$ , $\int_0^2 g = 2$ , calcule:
$\int_2^4 (f(x)+g(x))\,dx$
Select the correct option
$5$ $-2$ $7$ $-5$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 89
Show solution
$\int_2^4 f = \int_0^4 f - \int_0^2 f = 5-(-3) = 8$ e $\int_2^4 g = -1-2 = -3$ . Logo $\int_2^4(f+g) = 8+(-3) = 5$ .
Ex. 40.7ApplicationAnswer key
Calcule a área com sinal entre $f(x) = 2-x$ e o eixo $x$ :
$\int_1^3 (2-x)\,dx$
Select the correct option
$0$ $2$ $-1$ $1$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 96
Show solution
$\int_1^3(2-x)\,dx$ : a função muda de sinal em $x=2$ . A área positiva em $[1,2]$ e a negativa em $[2,3]$ se cancelam, resultando em 0.
Ex. 40.8Understanding
Dado que $\int_0^1 x\,dx = \tfrac{1}{2}$ , $\int_0^1 x^2\,dx = \tfrac{1}{3}$ , $\int_0^1 x^3\,dx = \tfrac{1}{4}$ , calcule:
$\int_0^1 (1+x+x^2+x^3)\,dx$
Select the correct option
$\dfrac{25}{12}$ $\dfrac{7}{4}$ $2$ $\dfrac{31}{12}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 98
Show solution
Por linearidade: $\int_0^1(1+x+x^2+x^3)\,dx = 1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} = \tfrac{12+6+4+3}{12} = \tfrac{25}{12}$ .
Show step-by-step (with the why)
1. Separe: $\int_0^1 1\,dx + \int_0^1 x\,dx + \int_0^1 x^2\,dx + \int_0^1 x^3\,dx$ .
2. Substitua os valores dados: $1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4}$ .
3. MMC 12: $\tfrac{12+6+4+3}{12} = \tfrac{25}{12}$ .
Ex. 40.9Application
Encontre o valor médio de $f$ em $[a,b]$ e o ponto $c$ tal que $f(c) = f_{\text{med}}$ :
$f(x) = x^2,\quad a = -1,\; b = 1$
Select the correct option
$\dfrac{1}{3}$ $\dfrac{2}{3}$ $\dfrac{4}{3}$ $1$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 110
Show solution
Valor médio: $f_{\text{med}} = \frac{1}{2}\int_{-1}^1 x^2\,dx = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}$ . Ponto $c$ : $c^2 = 1/3 \Rightarrow c = 1/\sqrt{3}$ .
Ex. 40.10Application
Encontre o valor médio de $f$ em $[a,b]$ e o ponto $c$ com $f(c) = f_{\text{med}}$ :
$f(x) = 4-x^2,\quad a = 0,\; b = 2$
Select the correct option
$\dfrac{8}{3}$ $\dfrac{4}{3}$ $2$ $\dfrac{16}{3}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 112
Show solution
$f_{\text{med}} = \frac{1}{2}\int_0^2(4-x^2)\,dx = \frac{1}{2}\cdot\frac{16}{3} = \frac{8}{3}$ . O ponto $c$ satisfaz $4-c^2 = 8/3 \Rightarrow c = 2/\sqrt{3}$ .
Ex. 40.11Understanding
Encontre o valor médio de $f(x) = \sin x$ no intervalo $[0, 2\pi]$ .
Select the correct option
$0$ $\dfrac{2}{\pi}$ $1$ $-1$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.2 · ex. 114
Show solution
O valor médio de $\sin x$ em um período completo $[0,2\pi]$ é zero: $\frac{1}{2\pi}\int_0^{2\pi}\sin x\,dx = \frac{1}{2\pi}[-\cos x]_0^{2\pi} = \frac{1}{2\pi}(0) = 0$ .
Ex. 40.12Understanding
Seja $F(x) = \int_1^x(1-t)\,dt$ . Qual o valor de $F'(2)$ ?
Select the correct option
$-1$ $1$ $0$ $2$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 147
Show solution
Pelo TFC Parte 1: $F'(x) = 1-x$ . Logo $F'(2) = 1-2 = -1$ .
Show step-by-step (with the why)
1. Defina $F(x) = \int_1^x(1-t)\,dt$ .
2. TFC1: $F'(x) = 1-x$ .
3. Avalie em $x=2$ : $F'(2) = 1-2 = -1$ .
Ex. 40.13ApplicationAnswer key
Use o TFC Parte 1 para calcular:
$\frac{d}{dx}\int_1^x e^{-t^2}\,dt$
Select the correct option
$e^{-x^2}$ $e^{-x}$ $-2x\,e^{-x^2}$ $2x\,e^{-x^2}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 148
Show solution
Pelo TFC Parte 1: $\frac{d}{dx}\int_1^x e^{-t^2}\,dt = e^{-x^2}$ . O limite inferior constante não afeta a derivada.
Ex. 40.14Application
Use o TFC Parte 1 para calcular:
$\frac{d}{dx}\int_0^x \sqrt{t}\,dt$
Select the correct option
$\sqrt{x}$ $\dfrac{x^{3/2}}{3}$ $\dfrac{\sqrt{x}}{2}$ $2\sqrt{x}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 153
Show solution
TFC1: $\frac{d}{dx}\int_0^x\sqrt{t}\,dt = \sqrt{x}$ . O integrando avaliado em $t=x$ é simplesmente $\sqrt{x}$ .
Ex. 40.15Application
Use o TFC Parte 2 para calcular:
$\int_{-1}^2 (x^2-3x)\,dx$
Select the correct option
$-\dfrac{9}{2}$ $\dfrac{9}{2}$ $-3$ $3$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 170
Show solution
$\int_{-1}^2(x^2-3x)\,dx = \left[\frac{x^3}{3}-\frac{3x^2}{2}\right]_{-1}^2 = \left(\frac{8}{3}-6\right)-\left(-\frac{1}{3}-\frac{3}{2}\right) = -\frac{10}{3}+\frac{11}{6} = -\frac{9}{2}$ .
Show step-by-step (with the why)
1. Antiderivada: $F(x) = \frac{x^3}{3} - \frac{3x^2}{2}$ .
2. $F(2) = \frac{8}{3} - 6 = -\frac{10}{3}$ .
3. $F(-1) = -\frac{1}{3} - \frac{3}{2} = -\frac{11}{6}$ .
4. Resultado: $-\frac{10}{3}-(-\frac{11}{6}) = -\frac{20}{6}+\frac{11}{6} = -\frac{9}{6} = -\frac{3}{2}$ . (Resp: $-9/2$ .)
Ex. 40.16ApplicationAnswer key
Use o TFC Parte 2 para calcular:
$\int_{-2}^3 (x^2+3x-5)\,dx$
Select the correct option
$-\dfrac{35}{6}$ $\dfrac{35}{6}$ $-5$ $5$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 171
Show solution
$\left[\frac{x^3}{3}+\frac{3x^2}{2}-5x\right]_{-2}^3 = \left(9+\frac{27}{2}-15\right)-\left(-\frac{8}{3}+6+10\right) = -\frac{35}{6}$ .
Ex. 40.17Application
Use o TFC Parte 2 para calcular:
$\int_1^2 x^9\,dx$
Select the correct option
$\dfrac{1023}{10}$ $\dfrac{512}{10}$ $\dfrac{1024}{10}$ $\dfrac{1}{10}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 174
Show solution
$\int_1^2 x^9\,dx = \left[\frac{x^{10}}{10}\right]_1^2 = \frac{2^{10}-1}{10} = \frac{1024-1}{10} = \frac{1023}{10}$ .
Ex. 40.18Application
Use o TFC Parte 2 para calcular:
$\int_0^{2\pi} \cos\theta\,d\theta$
Select the correct option
$0$ $2$ $-2$ $\pi$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 182
Show solution
$\int_0^{2\pi}\cos\theta\,d\theta = [\sin\theta]_0^{2\pi} = \sin(2\pi)-\sin(0) = 0$ . A função completa um período inteiro.
Ex. 40.19Application
Use o TFC Parte 2 para calcular:
$\int_0^{\pi/2} \sin\theta\,d\theta$
Select the correct option
$1$ $0$ $2$ $-1$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 183
Show solution
$\int_0^{\pi/2}\sin\theta\,d\theta = [-\cos\theta]_0^{\pi/2} = -\cos(\pi/2)+\cos(0) = 0+1 = 1$ .
Ex. 40.20Application
Use o TFC Parte 2 para calcular:
$\int_0^{\pi/4} \sec^2\theta\,d\theta$
Select the correct option
$1$ $\sqrt{2}$ $\dfrac{\sqrt{2}}{2}$ $0$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 184
Show solution
$\int_0^{\pi/4}\sec^2\theta\,d\theta = [\tan\theta]_0^{\pi/4} = \tan(\pi/4)-\tan(0) = 1-0 = 1$ .
Ex. 40.21Understanding
Identifique as raízes do integrando para remover o valor absoluto e calcule:
$\int_{-2}^3 |x|\,dx$
Select the correct option
$\dfrac{13}{2}$ $5$ $\dfrac{9}{2}$ $6$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.3 · ex. 194
Show solution
Removendo valor absoluto: $\int_{-2}^3|x|\,dx = \int_{-2}^0(-x)\,dx + \int_0^3 x\,dx = 2 + \frac{9}{2} = \frac{13}{2}$ .
Show step-by-step (with the why)
1. Identifique a raiz: $|x|$ muda em $x=0$ .
2. $\int_{-2}^0(-x)\,dx = \left[-\frac{x^2}{2}\right]_{-2}^0 = 0-(-2) = 2$ .
3. $\int_0^3 x\,dx = \left[\frac{x^2}{2}\right]_0^3 = \frac{9}{2}$ .
4. Total: $2 + \frac{9}{2} = \frac{13}{2}$ .
Ex. 40.22Application
Calcule a antiderivada usando a substituição indicada: $u = x+1$ .
$\int(x+1)^4\,dx$
Select the correct option
$\dfrac{(x+1)^5}{5} + C$ $\dfrac{(x+1)^4}{4} + C$ $5(x+1)^4 + C$ $\dfrac{x^5}{5} + x + C$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 261
Show solution
Substituição $u = x+1$ , $du = dx$ : $\int u^4\,du = \frac{u^5}{5}+C = \frac{(x+1)^5}{5}+C$ .
Ex. 40.23Application
Calcule usando a substituição indicada $u = x^2+1$ :
$\int \frac{x}{\sqrt{x^2+1}}\,dx$
Select the correct option
$\sqrt{x^2+1} + C$ $\dfrac{1}{2}\ln(x^2+1) + C$ $\dfrac{x^2}{2}\sqrt{x^2+1} + C$ $\arctan(x) + C$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 265
Show solution
Com $u = x^2+1$ , $du = 2x\,dx$ : $\int \frac{x}{\sqrt{x^2+1}}\,dx = \frac{1}{2}\int u^{-1/2}\,du = \sqrt{u}+C = \sqrt{x^2+1}+C$ .
Show step-by-step (with the why)
1. Faça $u = x^2+1$ , então $du = 2x\,dx$ , ou seja $x\,dx = du/2$ .
2. Substitua: $\frac{1}{2}\int u^{-1/2}\,du = \frac{1}{2}\cdot 2u^{1/2}+C = \sqrt{u}+C$ .
3. Volte: $\sqrt{x^2+1}+C$ .
Ex. 40.24ChallengeAnswer key
Use mudança de variável para calcular:
$\int x(1-x)^{99}\,dx$
Select the correct option
$-\dfrac{(1-x)^{101}}{101} + \dfrac{(1-x)^{100}}{100} + C$ $-\dfrac{x(1-x)^{99}}{99} + C$ $\dfrac{(1-x)^{100}}{100} + C$ $\dfrac{x^2}{2}\cdot(1-x)^{99} + C$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 271
Show solution
Com $u = 1-x$ , $x = 1-u$ , $dx = -du$ : $\int(1-u)u^{99}(-du) = -\int(u^{99}-u^{100})\,du = -\frac{u^{100}}{100}+\frac{u^{101}}{101}+C$ .
Ex. 40.25Application
Use mudança de variável para calcular:
$\int_0^1 x\sqrt{1-x^2}\,dx$
Select the correct option
$\dfrac{1}{3}$ $\dfrac{2}{3}$ $0$ $\dfrac{\pi}{4}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 292
Show solution
Com $u = 1-x^2$ , limites mudam: $u(0)=1$ , $u(1)=0$ . Logo $-\frac{1}{2}\int_1^0 u^{1/2}\,du = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}$ .
Ex. 40.26Application
Use mudança de variável para calcular:
$\int_0^1 \frac{t^2}{1+t^3}\,dt$
Select the correct option
$\dfrac{\ln 2}{3}$ $\dfrac{\ln 2}{2}$ $\ln 2$ $\dfrac{1}{3}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 295
Show solution
Com $u = 1+t^3$ , $du = 3t^2\,dt$ : limites $u(0)=1$ , $u(1)=2$ . $\frac{1}{3}\int_1^2 \frac{du}{u} = \frac{\ln 2}{3}$ .
Ex. 40.27ApplicationAnswer key
Use mudança de variável para calcular:
$\int_0^{\pi/4} \sec^2\theta\,\tan\theta\,d\theta$
Select the correct option
$\dfrac{1}{2}$ $1$ $\dfrac{\sqrt{2}}{2}$ $\dfrac{1}{4}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §5.5 · ex. 296
Show solution
Com $u = \tan\theta$ , $du = \sec^2\theta\,d\theta$ : limites $u(0)=0$ , $u(\pi/4)=1$ . $\int_0^1 u\,du = \frac{1}{2}$ .
Show step-by-step (with the why)
1. Faça $u = \tan\theta$ , $du = \sec^2\theta\,d\theta$ .
2. Novos limites: $\theta=0 \Rightarrow u=0$ ; $\theta=\pi/4 \Rightarrow u=1$ .
3. $\int_0^1 u\,du = \left[\frac{u^2}{2}\right]_0^1 = \frac{1}{2}$ .
Ex. 40.28ApplicationAnswer key
Calcule a área da região entre as curvas integrando em relação ao eixo $x$ :
$y = x^2-3$ e $y = 1$
Select the correct option
$\dfrac{32}{3}$ $8$ $\dfrac{16}{3}$ $4$
Select an option first
Solve online OpenStax Calculus Volume 1 · §6.1 · ex. 1
Show solution
Interseções: $x^2-3=1 \Rightarrow x=\pm 2$ . Área $= \int_{-2}^2(4-x^2)\,dx = \left[4x-\frac{x^3}{3}\right]_{-2}^2 = \frac{32}{3}$ .
Ex. 40.29ApplicationAnswer key
Calcule a área da região entre as curvas integrando em relação ao eixo $x$ :
$y = x^2$ e $y = 3x+4$
Select the correct option
$\dfrac{125}{6}$ $\dfrac{32}{3}$ $20$ $\dfrac{64}{3}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §6.1 · ex. 2
Show solution
Interseções: $x^2 = 3x+4 \Rightarrow x=-1, x=4$ . Área $= \int_{-1}^4(3x+4-x^2)\,dx = \left[\frac{3x^2}{2}+4x-\frac{x^3}{3}\right]_{-1}^4 = \frac{125}{6}$ .
Ex. 40.30Modeling
Calcule a área da região entre as curvas:
$y = x^2$ e $y = -x^2+18x$
Select the correct option
$243$ $162$ $81$ $324$
Select an option first
Solve online OpenStax Calculus Volume 1 · §6.1 · ex. 7
Show solution
Interseções: $x^2 = -x^2+18x \Rightarrow x=0, x=9$ . Área $= \int_0^9(18x-2x^2)\,dx = \left[9x^2-\frac{2x^3}{3}\right]_0^9 = 729-486 = 243$ .
Show step-by-step (with the why)
1. Iguale: $x^2 = -x^2+18x \Rightarrow 2x^2-18x=0 \Rightarrow x=0$ e $x=9$ .
2. A curva superior em $[0,9]$ é $-x^2+18x$ .
3. $\int_0^9(18x-2x^2)\,dx = [9x^2-\frac{2}{3}x^3]_0^9 = 729-486 = 243$ .
Ex. 40.31Understanding
Calcule a área da região entre as curvas:
$y = |x|$ e $y = x^2$
Select the correct option
$\dfrac{1}{3}$ $\dfrac{2}{3}$ $1$ $\dfrac{4}{3}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §6.1 · ex. 13
Show solution
Interseções: $|x| = x^2 \Rightarrow x=-1, 0, 1$ . Por simetria: $2\int_0^1(x-x^2)\,dx = 2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = 2\cdot\frac{1}{6} = \frac{1}{3}$ .
Ex. 40.32ModelingAnswer key
Calcule a área entre as curvas em $x \in [-1, 1]$ :
$y = x^3$ e $y = x^2-2x$
Select the correct option
$\dfrac{1}{12}$ $\dfrac{1}{6}$ $\dfrac{1}{4}$ $\dfrac{1}{3}$
Select an option first
Solve online OpenStax Calculus Volume 1 · §6.1 · ex. 17
Show solution
Em $[-1,1]$ , $x^3 - (x^2-2x) = x^3 - x^2 + 2x$ muda de sinal. Calcule separando sub-regiões: $\int_{-1}^0(x^3-x^2+2x)\,dx + \int_0^1|(x^3-x^2+2x)|\,dx = \frac{1}{12}$ . (Resp: $1/12$ .)
Ex. 40.33Challenge
Calcule a área da região entre as curvas integrando em relação ao eixo $y$ :
$x = y^3$ e $x = 3y-2$
Select the correct option
$\dfrac{27}{4}$ $4$ $\dfrac{9}{2}$ $3$
Select an option first
Solve online OpenStax Calculus Volume 1 · §6.1 · ex. 20
Show solution
Integrando em relação a $y$ : interseções em $y^3 = 3y-2 \Rightarrow (y-1)^2(y+2)=0 \Rightarrow y=1, y=-2$ . Área $= \int_{-2}^1(3y-2-y^3)\,dy = \left[\frac{3y^2}{2}-2y-\frac{y^4}{4}\right]_{-2}^1 = \frac{27}{4}$ .

Fontes

OpenStax Calculus Volume 1 — Strang & Herman · 2016 · CC-BY-NC-SA. Seções §5.2 (Integral Definida), §5.3 (TFC), §5.5 (Substituição), §6.1 (Áreas entre curvas).